Define the sequence the following way for some $x,y \geq 0$:
$$a_0=x,~~~~~~~b_0=y$$
$$a_{n+1}=\frac{a_n^2+b_n^2}{a_n+b_n},~~~~~~b_{n+1}=\frac{a_n+b_n}{2}$$
Obviously:
$$a_n \geq b_n,~~~~n \geq 1$$
For convergence rate we have:
$$a_{n+1}-b_{n+1}=\frac{(a_n-b_n)^2}{2(a_n+b_n)} \tag{1}$$
We have a weak inequality:
$$\frac{(a_n-b_n)^2}{2(a_n+b_n)} \leq \frac{(a_n-b_n)^2}{2(a_n-b_n)}=\frac{a_n-b_n}{2}$$
So our convergence rate is at least linear:
$$\frac{a_{n+1}-b_{n+1}}{a_n-b_n} \leq \frac{1}{2} \tag{2}$$
But shouldn't $(1)$ imply faster (quadratic) convergence?
(I know we need to subtract the limit to find the convergence, but I don't know the closed form, see below).
Now for the limit.
We have the following relations:
$$a_{n+1}b_{n+1}=\frac{a_n^2+b_n^2}{2} \geq a_nb_n$$
$$a_{n+1}b_{n+1}-a_nb_n=\frac{(a_n-b_n)^2}{2}$$
Can we find the limit of this sequence in closed form?
What is the true rate of convergence for this sequence?